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Electric field and electric potential doubt

  1. May 26, 2010 #1
    Hi =D

    I know that the electric potential in any given point (considering only a point charge) in a region is defined to be:

    [tex]\varphi = \frac{q}{4\pi \varepsilon_{0} d}[/tex]

    I know too that the electric potential generally can be defined as:

    [tex]\varphi = - \int_{P1}^{P2} \overrightarrow{E} \cdot \overrightarrow{ds}[/tex]

    The line integral of any path between P1 and P2 blah blah blah, you already know...

    But what I really want to know is: how do derive the first simplified equation from this second general equation?

    I got some progress but I still having trouble:

    (Considering a region in space with only a point charge).

    Because of symmetry we know that through any point I choose passes exact one electric field line and this line passes through the point charge too. So the electric field line is always parallel to the segment that connects point charge and the point I choose. Thus the line integral with the dot product inside can be replaced by the electric field strength multiplied by the distance between the point charge and the point I choose. Is that right?

    So, what about the minus before the integral? Why it's in the first equation? I'm I thinking wrong about something? If someone could help me I would be grateful...

    If someone has an analogy to why the electric potential decreases with the distance (and not the square of the distance) I would be grateful too...

    Thank you,
    Rafael Andreatta
  2. jcsd
  3. May 26, 2010 #2
    Since you are relating the potential to the potential difference you need to take one bound of the integral to be infinity(zero voltage reference) and the other to be the point d. The potential is a measure of the amount of energy it it takes to move from infinity to some point close to the point charge in the electric field generated by the point charge in this case. So the bounds are inf -> d (This is in contrast to the electric field which points away from its source).
    It's also dL(vector) not dS(vector), it could be dS but you need to take stokes theorem or something and it would look different. dL(vector) = dr*ar(vector) since it moves out radially.

    E(vector) = [q / (4*pi*eo*r^2)]*ar(vector) as you said this is inverse square law(also the reason why the potential is not inverse square law is because you integrate over a line and not measure at a single point).

    Yea so dot them and integrate and it should look the same.
  4. May 30, 2010 #3
    Thank you for the reply. But I didn't understand really your post...

    Why it should be dL and not dS (what is the dL that you're talking about?). What is the ar vector that you are talking about?

    Could you explain it more clearly, please?

    Thank you
  5. May 30, 2010 #4
    The reason why it is dL and not dS is by definition of the equation you need to use, you may have selected the wrong one or an out of context one. Admittedly dL is taken over a closed loop, so it does enclose a surface, but in this configuration of the question where you are moving across the electric field generated by a point charge, the field is conservative and you only need to take dL and integrate over a line and not a loop. If you wish to apply green's(green's, stoke's, and divergence are similar sorry if i mixed them up) theorem and integrate over the surface, the inside of the equation where E is would not be the same, it would have some operation altering it, therefore do not be bogged down in this emthod but instead concentrate on the more pertinent equation and use dL.

    ar or just r is the radial unit vector under the assumption the point charge is in 3 space.
  6. May 31, 2010 #5
    Thank you for the replies again.

    But I can't really understand you... what equation did I select wrong? The electric potential one? Isn't the electric potential defined as [tex]
    \varphi = - \int_{P1}^{P2} \overrightarrow{E} \cdot \overrightarrow{ds}

    What do closed loops have to do with my question?

    If it doesn't bother you too much, could you try to explain me that with doing the step-by-step equations? And if my question has really something to do with green's theorem, could you explain me what does this theorem states?

    Thank you again...
  7. May 31, 2010 #6
    That is not the definition of the electric potential difference. If you look at what you wrote you see you are integrating across a surface, dS right? therefore it would have to be a double integral, so that equation itself is not even mathematically correct, please look up the correct equation rather then defending that one.

    If you take a look at my original reponse I set up the integral for you to solve, and define the variables. It is no longer electromagnetics just some mathematics, try to set up what I wrote to set you on the right course.
  8. May 31, 2010 #7
    I believe these are line integrals, and not loop integrals (loop integrals usually use o-integrals, where there is a circle drawn over the summation to indicate the loopiness).

    ds or dl is just simply a naming convention, what they actually are is context sensitive.

    I assume it is safe, here, to assume ds is just one dimensional, so I will use ds=d(r).

    The electric field of a point particle is easy enough to look up, it is

    Then we can just integrate:

    [tex]\varphi =-\int_{\infty}^{d}\frac{q}{4\pi\epsilon_{0}r^{2}}dr=-\frac{q}{4\pi\epsilon_{0}}\int_{\infty}^{d}\frac{1}{r^{2}}dr=\left[\frac{q}{4\pi\epsilon_{0}r}\right]_{\infty}^{d}=\frac{q}{4\pi\epsilon_{0}d}-\frac{q}{4\pi\epsilon_{0}\infty}=\frac{q}{4\pi\epsilon_{0}d}[/tex]

    Let me know if you are confused as to WHY I have selected the bounds of my integral the way I have.

    I also want to make certain for you that, GENERALLY SPEAKING:
    [tex]\frac{q}{4\pi\epsilon_{0}d}\neq\int_{P1}^{P2}\vec{E}\cdot d\vec{s}[/tex]

    It's ONLY true when,
    [tex]\frac{q}{4\pi\epsilon_{0}d}=\int_{\infty}^{d}\vec{E}\cdot d\vec{s}[/tex]
    Last edited: May 31, 2010
  9. Jun 1, 2010 #8
    Thank you very much for the response, you solved my doubt...

    Just a little question (I think it's not physics but just mathematics, but I won't start a new thread on math foruns just for that): when you divide something by +inf could you consider it zero? Is that really defined in mathematics (division by infinity)? Or you just can do that in this case because you are taking that in the limit?

    Thank you again...
  10. Jun 1, 2010 #9
    Good question, actually, integrals are not defined at infinity and so we use an improper integral to calculate the integral.

    In Calculus 2 (or something) you learn that

    Is this a satisfactory enough answer?
  11. Jun 1, 2010 #10
    Yes, thank you very much...
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